python从一个函数调用另一个函数
python从一个函数调用另一个函数
提问于 2019-02-20 15:48:39
我正在使用python django DRF .Im创建一个序列化程序，创建一个单独的助手类，并尝试覆盖create方法。
我的助手类：
class WorkFlowHelper:
def assign_level_permissions(self,workflow_level,level):
for permission in level['permissions']:
workflow_permission_obj = WorkflowPermission.objects.get(short_name=permission['short_name'])
workflow_level.permissions.add(workflow_permission_obj)
def create_levels(self,workflow,levels):
for level in levels:
workflow_level = WorkflowLevel()
workflow_level.workflow=workflow,
workflow_level.level = level['level']
workflow_level.operation=level['operation']
workflow_level.save()
workflow.levels.add(workflow_level)
self.assign_level_permissions(workflow_level,level)
def create_categories(self,workflow, categories):
for category in categories:
workflow_category_obj = WorkflowCategory.objects.get(short_name=category['short_name'])
workflow.categories.add(workflow_category_obj)
def create(self,name, description,tenant, levels, categories):
workflow = Workflow.objects.create(name=name, description=description, tenant=tenant)
self.create_levels(workflow,levels)
self.create_categories(workflow,categories)
workflow.save()
return workflow在序列化程序中，我必须调用create方法，如下所示：
def create(self, validated_data):
name=validated_data['name']
description=validated_data.get('description'),
tenant=self.context['request'].user.tenant
levels = self.initial_data['levels']
categories = self.initial_data['categories']
helper = WorkFlowHelper()
helper.create(name,description,tenant,levels,categories)在helper类中，im从同一个类中的一个函数调用其他函数。我对如何在函数中传入call.Now有疑问，我不能保存模型，一切都能正常工作，但就在post之后，它显示的create()没有返回object instanace AssertionError..but它正在保存模型并正常工作
回答 2
Stack Overflow用户
回答已采纳
发布于 2019-02-20 16:32:09
这是可行的：
class WorkFlowHelper:
def assign_level_permissions(self,workflow_level,level):
for permission in level['permissions']:
workflow_permission_obj = WorkflowPermission.objects.get(short_name=permission['short_name'])
workflow_level.permissions.add(workflow_permission_obj)
def create_levels(self,workflow,levels):
for level in levels:
workflow_level = WorkflowLevel()
workflow_level.workflow=workflow,
workflow_level.level = level['level']
workflow_level.operation=level['operation']
workflow_level.save()
workflow.levels.add(workflow_level)
self.assign_level_permissions(workflow_level,level)
def create_categories(self,workflow, categories):
for category in categories:
workflow_category_obj = WorkflowCategory.objects.get(short_name=category['short_name'])
workflow.categories.add(workflow_category_obj)
def create(self,name, description,tenant, levels, categories):
workflow = Workflow.objects.create(name=name, description=description, tenant=tenant)
self.create_levels(workflow,levels)
self.create_categories(workflow,categories)
workflow.save()
return workflow在序列化程序中：
def create(self, validated_data):
name=validated_data['name']
description=validated_data.get('description'),
tenant=self.context['request'].user.tenant
levels = self.initial_data['levels']
categories = self.initial_data['categories']
helper = WorkFlowHelper()
workflow = helper.create(name,description,tenant,levels,categories)
return workflowStack Overflow用户
发布于 2019-02-20 15:55:42
self是对对象本身的引用。使用self作为函数的第一个参数，您可以创建实例方法。因此，如果要使用instance方法，则需要初始化该类的对象，然后使用该对象调用函数。例如，在这里：
helper = WorkFlowHelper() # create a `WorkFlowHelper` object
helper.create(creates,name,description,tenant,levels,categories)
# here you don't need to pass `self`, because it is referencing to it to the helper object作为参考，您可以查看here。
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原文链接：
https://stackoverflow.com/questions/54781263
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